# ΔABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2. Given: ABC is an isosceles triangle right angled at C.

Let AC = BC

In ∆ACB, using Pythagoras theorem, we have

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AC)2 + (BC)2 = (AB)2

(AC)2 + (AC)2 = (AB)2

[ABC is an isosceles triangle, AC =BC]

2(AC)2 = (AB)2

Hence Proved

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